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Journal of Function Spaces
Volume 2015 (2015), Article ID 865069, 9 pages
http://dx.doi.org/10.1155/2015/865069
Research Article

The Singular Points of Analytic Functions with Finite -Order Defined by Laplace-Stieltjes Transformations

1Department of Informatics and Engineering, Jingdezhen Ceramic Institute, Jingdezhen, Jiangxi 333403, China
2Beijing Key Laboratory of Information Service Engineering, Department of General Education, Beijing Union University, No. 97 Bei Si Huan Dong Road, Chaoyang District, Beijing 100101, China

Received 19 April 2014; Accepted 18 July 2014

Academic Editor: Seppo Hassi

Copyright © 2015 Hong-Yan Xu and Zu-Xing Xuan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study the singular points of analytic functions defined by Laplace-Stieltjes transformations which converge on the right half plane, by introducing the concept of -order functions. We also confirm the existence of the finite -order Borel points of such functions and obtained the extension of the finite -order Borel point of two analytic functions defined by two Laplace-Stieltjes transformations convergent on the right half plane. The main results of this paper are improvement of some theorems given by Shang and Gao.

1. Introduction

For Laplace-Stieltjes transforms where is a bounded variation on any interval and and are real variables. We choose a sequence , which satisfies the following conditions: where

Remark 1. Dirichlet series was regarded as a special example of Laplace-Stieltjes transformations; a number of articles have focused on the growth and the value distribution of analytic functions defined by Dirichlet series; see [13] for some recent results.

In 1963, Yu [4] proved the Valiron-Knopp-Bohr formula of the associated abscissas of bounded convergence, absolute convergence, and uniform convergence of Laplace-Stieltjes.

Theorem A. Suppose that Laplace-Stieltjes transformations (1) satisfy the first formula of (3) and ; then where is called the abscissa of uniform convergence of .

By (3), (4), and Theorem A, one can get that ; that is, is analytic in the right half plane. Set

Remark 2. From (4), for any , we have This shows that exists.

In the past few decades, many people studied some problems of analytic functions defined by Laplace-Stieltjes transformations and obtained a number of interesting and important results (including [511]). In those papers, there are about two methods to control the growth of the maximum modulus or the maximum term : one method is to replace the denominator in the definition of growth order by using the technique of type function (see [4, 1214]) and the other method is to take multiple logarithm to or in the definition of growth order (see [15, 16]). For the second method, as the logarithm function is a special function, a question rises naturally: whether we can find a relatively general function to replace the logarithm function to control the maximum growth rate.

In this paper, we investigate the above question and give a positive answer to this question. Moreover, we confirm the existence of singular points for these functions by applying the main results of our paper. To do this, we introduce a completely new technique based on the concept of which is different with the type function and more general than logarithm function and obtain the main theorems as follows.

Theorem 3. If the Laplace-Stieltjes transformation of infinite order has finite -order and the sequence (2) satisfies (3) and (4), then one has

Theorem 4. If the Laplace-Stieltjes transformation of infinite order and the sequence (2) satisfy (3) and (4), then one has where .

Remark 5. The definitions of -order and the function in Theorems 3 and 4   will be introduced in Section 2.

From Theorem 4, we further investigate the value distribution of analytic functions with finite -order represented by Laplace-Stieltjes transformations convergent on the right half plane and obtain the following theorems.

Theorem 6. Suppose that the sequence (2) satisfies (3) and (4) and Laplace-Stieltjes transformation is of infinite order. Let , where is a creasing function, and, for any positive number and , satisfies and then is the -point of with -order ; that is, for any , the inequality holds for any , except for one exception, where is the counting function of distinct zero of the function in the strip .

Theorem 7. Suppose that the sequence (2) satisfies (3) and (4) and Laplace-Stieltjes transformation is of infinite order. Let , where is continuous function on , , is a positive real number, and then is the -point of with -order ; that is, for any , the inequality holds for any , except for one exception, where is the counting function of distinct zeros of the function in the strip .

The structure of this paper is as follows. In Section 2, we introduce the concept of -order and give the proofs of Theorems 3 and 4. Section 3 is devoted to proving Theorems 6 and 7.

Remark 8. From Theorems 37, we assume that the -order of is finite; that is, . For , we have studied the value distribution of analytic functions defined by Laplace-Stieltjes transformations which converge on the right half plane and obtained some results (see [17]).

2. Proofs of Theorems 3 and 4

We first introduce the concept of -order of such functions as follows.

Definition 9 (see [18]). If the Laplace-Stieltjes transform satisfies (the sequence (2) satisfies (3) and (4)) and then is called a Laplace-Stieltjes transform of infinite order.

To control the growth of the molecule or in the definition of order, many mathematicians proposed the type functions to enlarge the growth of the denominator or (see [4, 6, 11, 13, 14]). In this paper, we will investigate the growth of Laplace-Stieltjes transform of infinite order by using a class of functions to reduce the growth of or which is different with the previous form. Thus, we should give the definition of the new function as follows.

Let be the class of all functions which satisfies the following conditions:(i) is defined on , is positive, strictly increasing, and differential, and tends to as ;(ii) as .

Definition 10. If the Laplace-Stieltjes transformation of infinite order satisfies where , then is called the -order of the Laplace-Stieltjes transform .

Remark 11. In particular, if we take , where and , -order is -order of Laplace-Stieltjes transformations with infinite order.

Remark 12. In addition, -order is more precise than -order. In fact, for ≥2 being a positive integer, we can find out function and a positive real function satisfying For example, letting , where is a finite positive real constant and .

The following lemma is very important to study the growth of analytic functions represented by Laplace-Stieltjes transforms convergent on the right half plane, which show the relation between and of such functions.

Lemma 13 (see [7, 11]). If the abscissa of uniform convergence of Laplace-Stieltjes transformation and the sequence (2) satisfy (3), then, for any given and for sufficiently reaching 0, one has where is a constant depending on , (3), and

2.1. The Proof of Theorem 3

Firstly, for any constant , we will prove that Without loss of generality, we suppose that . From the Laplace-Stieltjes transformation with infinite order and Lemma 13, we have . Then, from the Cauchy mean value theorem, there exists satisfying that is, From as and the above equality, we can easily get (20).

Thus, from (20) and Lemma 13, we can prove the conclusion of Theorem 3 easily.

2.2. The Proof of Theorem 4

: Suppose that

We consider two steps in this case as follows.

Step 1. We will prove that . From (23) and , there exists a positive sequence ; for any , From (18) and (23), we have Setting and taking , , then, from (24) and (25), we have that is, From , we have Thus, from (27), (28) and being strictly increasing function, we have Since , from the Cauchy mean value theorem, there exists satisfying that is, Since as , from (23) and (31), we can get

Step 2. We will prove that .
From and (23), there exists a positive integer ; for any , we have From (3), for any and the above , there exists a positive integer satisfying where and are two reciprocally inverse functions. Set , and then From the above inequality and (34), we have
Next, we will prove that By using the same argument as in (31), we can get that Since is infinite order and finite -order, from the definition of , we can get the following equality easily: From (38), (39) and , we have Thus, from (40), we can deduce From (36), we can get that That is, Hence, we have
Since and as , then there exists a positive integer satisfying Take From (45) and being a strictly increasing function, we can get that . Then, from (33), (45), (47), and , we have For any and any , we have Thus, from (33), (34), (45)–(47), and the above inequality, we have where is the sum of finite items of the series . Hence, from the above inequality, we have that is, Then, by using the same argument as in (31), we can deduce From Steps 1 and 2, the sufficiency of the theorem is completed.

By using the same argument as in the above proof of the theorem, we can prove the necessity of the theorem easily.

Thus, we complete the proof of Theorem 4.

3. Proofs of Theorems 6 and 7

Similar to the definition of -order of Laplace-Stieltjes transformations in the right half plane, we will give the definition of -order of Laplace-Stieltjes transformations in the level half-strip as follows.

Definition 14. Let be the analytic function with infinite order represented by Laplace-Stieltjes transformations convergent on the right half plane; set , where is a real number and is a positive number. Let and where ; then is called the -order of in the level half-strip .

To prove Theorems 6 and 7, we need some lemmas as follows.

Lemma 15. If the Laplace-Stieltjes transformation of infinite order and the sequence (2) satisfy (3) and (4) and , where is a creasing function, and, for any positive number and , satisfies then, for any , one has

Proof. We will prove this lemma by using the similar argument as in [11]. From the assumptions of Lemma 15, for any , Thus, we have
Since is an analytic function with infinite order, from the above inequality and the definition of -order, we can get the conclusion of Lemma 15 easily.

Lemma 16 (see [11, Lemma 2.4]). Let Then, one has the following:(i)this mapping maps the horizontal half-strip to the unit disc , and its inverse mapping is (ii);(iii);(iv).

Definition 17 (see [19, 20]). Let be a meromorphic function in and . Then is called the (upper) index of inadmissibility of . If , is called nonadmissible; if , is called admissible.

Lemma 18 (see [20, Theorem 2]). Let be a meromorphic function in and , let be a positive integer, and let be pairwise distinct complex numbers. Then, for , ,

Remark 19. We can see that holds in Lemma 18 without a possible exception set when .

Remark 20. The term in Lemma 18 can enter the remainder when the function satisfies .

From Lemmas 3.3 and 3.4 in [21], we can get the following result for nonadmissible functions in the unit disc which is used in this paper.

Lemma 21 (see [19, page 282 (1.8)]). Let be analytic in the disc ; then where is the maximum modulus of in the disc .

3.1. The Proof of Theorem 6

From the sequence (2) satisfying (3) and (4), Laplace-Stieltjes transformation of infinite order, and , from Theorem 4, we have and, from Lemma 15 and (63), for any , we have Thus, we can get where and .

Set , where is stated as in Lemma 16; then, from Lemma 16, we have that is analytic in the unit disc and satisfies where . Therefore, from (66) and Lemma 16, we have From Lemma 21, by using the same argument as in (31), we can get From (67), (68), and Lemma 21, we can get that is an admissible function in and Then, from Lemma 18, we can get that there at most exists one except value satisfying that is, for any , the inequality holds for any , except for one exception, where is the counting function of zeros of the function in the strip .

Thus, we complete the proof of Theorem 6.

3.2. The Proof of Theorem 7

Since is a continuous function on , then we can get that is a function of bounded variation on . Set From the assumptions of Theorem 7, for any real number , we have that is, From the assumption of Theorem 7, by (74) and Theorem 4, we can get From (75), for any , we have where .

Set , where is stated as in Lemma 16; then, from Lemma 16, we have that is analytic in the unit disc and satisfies where . Therefore, from the above inequality and Lemma 16, we have Then, similar to the proof of Theorem 6, we can prove that, for any , the inequality holds for any , except for one exception, where is stated as in Theorem 7.

Thus, we complete the proof of Theorem 7.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (11301233 and 61202313), the Natural Science Foundation of Jiangxi Province in China (20132BAB211001), and the Foundation of Education Department of Jiangxi (GJJ14644) of China. The second author is supported in part by Beijing Natural Science Foundation (no. 1132013) and The Project of Construction of Innovative Teams and Teacher Career Development for Universities and Colleges Under Beijing Municipality (CIT and TCD20130513).

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